Mathematical Structure of uva12716 gcd xor Number Theory

The question gives you an N, asking you to calculate two numbers A and B, and a> = B <= N, yes, gcd (a, B) = a ^ B

The range of N is 3*10 ^ 7, which is scary. At first, I didn't dare to construct it, because even if the array to be constructed is too large, it's already 10 ^ 7, after thinking about the ^ operation for a long time, I didn't come up with anything here, so I had no way to construct it. I constructed two numbers according to the meaning of his question, in J, J is a multiple of I, so the maximum common divisor of J + I and I must be I, then (J + I) ^ I = I, the constructed result is satisfied, and then the GCD will be imitated to put them in an array for counting, so that preprocessing can be done.

After playing well, I ran another brute-force program to run the answer. The result was correct, but the transaction timed out, because the preprocessing at the beginning assigned the long type, there is a % operation in the Process of division, which will lead to a very slow operation, so you can change it to int.

#define  _CRT_SECURE_NO_WARNINGS/*#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<string>#include<math.h>using namespace std;#define IN freopen("c:\\Users\\nit\\desktop\\input.txt", "r", stdin)    #define OUT freopen("c:\\Users\\nit\\desktop\\output.txt", "w", stdout)   int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int main(){OUT;int ans[510],k=0;memset(ans,0,sizeof(ans));for(int i=1;i<500;i++){for(int b=1;b<=i;b++){for(int a=b;a<=i;a++){if((a^b)==gcd(a,b))ans[i]++;}}printf("%d\t",ans[i]);if(k%10==0)puts("");k++;}return 0;}*/#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<string>#include<math.h>using namespace std;//#define IN freopen("c:\\Users\\nit\\desktop\\input.txt", "r", stdin)    //#define OUT freopen("c:\\Users\\nit\\desktop\\outpu1t.txt", "w", stdout)   #define ll long long#define MAXN 30000000 + 5ll ans[MAXN];void init() {for(ll i = 1;i<MAXN;i++) {for(ll j = i;i + j < MAXN;j += i) {if( ((i + j)^j) == i) {int x = j;int y = i + j;for(;x > 0 && y > 0;) {int tmp = x%y;x = y;y = tmp;if((x + y) == i)ans[i + j]++;}}}}for(int i = 2;i<MAXN;i++)ans[i] += ans[i-1];}int main() {init();int t;scanf("%d",&t);int Case = 0;while(t--) {int n;scanf("%d",&n);printf("Case %d: %lld\n",++Case,ans[n]);}return 0;}